Properties of convergent sequences theorem help I am reviewing some material about sequences and I ran across the following theorem.
Theorem: Let $(x_{n})$ and $(y_{n})$ be two convergent sequences with limits $x$ and $y$, respectively. Then $(x_n + y_n)$ converges to $x+y$.
The proof that the text uses is as follows.
Proof:
Since $x_n$ and $y_n$ are convergent, given $\epsilon > 0$, $\exists N_1,N_2$ such that $x_n \rightarrow x$ and $y_n \rightarrow y$ for all $n \geq N_1$, $n \geq N_2$, respectively. Taking $N=\max\{N_1,N_2\}$, and with the use of the triangle inequality it follows that 
$$|(x_n+y_n)-(x+y)| \leq |x_n-x| + |y_n-y| \leq 2\epsilon.$$
The question that I have is with the last inequality. If $|x_n-x| + |y_n-y| \leq 2\epsilon$ is true then how does this imply the limit exists? 
Attempt at understanding
If $|x_n-x| + |y_n-y| \leq 2\epsilon$, then $2\epsilon$ is the largest neighborhood that contains the limit. Thus we can choose another epsilon $\epsilon'\neq\epsilon$ with $\epsilon'<\epsilon$ such that if the limit, in this case $(x+y) \in (x-2\epsilon,x+2\epsilon)$, then $(x+y) \in (x-\frac{\epsilon'}{2},x+\frac{\epsilon'}{2}) \in (x-2\epsilon,x+2\epsilon)$.
 A: $|(x_n+y_n)-(x+y)| \leq |x_n-x| + |y_n-y| \leq 2\epsilon.$ 
Here $2\epsilon$ is arbitrary. The above inequality tells us that  for any $2\epsilon-$ neighborhood of $(x+y)$, we can find a stage $N\in \Bbb N$ such that all terms of the sequence $(x_n+y_n)$ will lie in this neighborhood. 
This is precisely the condition which is required to show that a sequence converges to some point. Isn't it?
A: I would prove it in this way and this proof would stick to  the definition of limit:

If $x_n$ and $y_n$ are convergent, let $\lim_{n->\infty} x_n=L_1$ and $\lim_{n->\infty} y_n=L_2$.
Let $\epsilon$ be any given positive number.
Then for the positive number $\frac{\epsilon}{2}$, we can find a positive integer $N_1$ such that
$$|x_n-L_1|\lt \frac{\epsilon}{2}\text{ , for all }n>N_1\text{ ...(1)}$$
Again, for the positive number $\frac{\epsilon}{2}$, we can find a positive integer $N_2$ such that
$$|y_n-L_2|\lt \frac{\epsilon}{2}\text{ , for all }n>N_2\text{ ...(2)}$$
Take $N=\max\{N_1,N_2\}$, When $n>N$, both inequalities (1) and (2) hold. Hence
$$|(x_n+y_n)-(L_1+L_2)|=|(x_n-L_1)+(y_n-L_2)|\le |x_n-L_1|+|y_n-L_2|\lt \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$for all $n>N$.
Therefore $\lim_{n->\infty}(x_n+y_n)=L_1+L_2$

Perhaps this proof would help you understand it because it is derived directly from the definition of limit.
